orbital mechanics notes

5/28/2018 Orbital Mechanics notes

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Orbital Mechanics Course Notes

David J. WestpfahlProfessor of Astrophysics,

New Mexico Institute of Mining and Technology

March 31, 2011


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These are notes for a course in orbital mechanics catalogued as Aerospace

Engineering 313 at New Mexico Tech and Aerospace Engineering 362 atNew Mexico State University. This course uses the text Fundamentals ofAstrodynamics by R.R. Bate, D. D. Muller, and J. E. White, published byDover Publications, New York, copyright 1971. The notes do not follow thebook exclusively. Additional material is included when I believe that it isneeded for clarity, understanding, historical perspective, or personal whim.

We will cover the material recommended by the authors for a one-semestercourse: all of Chapter 1, sections 2.1 to 2.7 and 2.13 to 2.15 of Chapter 2,all of Chapter 3, sections 4.1 to 4.5 of Chapter 4, and as much of Chapters6, 7, and 8 as time allows.

PurposeThe purpose of this course is to provide an introduction to orbital me-

chanics. Students who complete the course successfully will be prepared toparticipate in basic space mission planning. By basic mission planning Imean the planning done with closed-form calculations and a calculator. Stu-dents will have to master additional material on numerical orbit calculationbefore they will be able to participate in detailed mission planning.

There is a lot of unfamiliar material to be mastered in this course. Thisis one field of human endeavor where engineering meets astronomy and ce-lestial mechanics, two fields not usually included in an engineering curricu-

lum. Much of the material that is familiar to students of those disciplineswill be unfamiliar to engineers. Students are probably already familiar withNewtons Laws and Newtonian gravity. These will be used to develop theparticular applications needed to describe orbits and orbital maneuvers.

Space missions are expensive and risky, especially if people or living ani-mals are sent into space. Thus, it is important to check and recheck calcula-tions and assumptions. Computer programs are subject to the imperfectionsof the humans who write them. This, it becomes necessary to develop phys-ical insight into orbit calculations to have a sense of when a programmingbug is leading to inaccurate answers. We will spend time developing physicalintuition and understanding what it is.

Notation

Well I remember being a student and being frustrated by notation usedby printers of textbooks that was impossible to write by hand at note-takingspeed, and not used by the professor, anyway. Thus, I have tried to use


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notation that is consistent with the text, but also within my abilities to write

on the board and within the abilities of students to write in their notes. Someexamples follow.

A scalar is written as an ordinary math symbol, as ina.

A vector is written with an arrow above, as inr.

A unit vector is written with a hat, as in r. If the unit vector is a basis vectorof a coordinate set its symbol is usually capitalized, as in I.

A matrix is written in a boldfaced capital letter and covered by a tilde, asin D. This is a compromise. The tilde is easy enough to write in notes oron the board, but boldface is not. The boldface is used to make the matrix

instantly recognizable in the notes, at the cost of inconsistency.The inverse of the same matrix is written as D1.

The transpose of a matrix is written in boldface with a tilde and a trailingsuperscript capital T, as in DT.

A triangle with vertices A, B, and C is named ABC. Its line segments arenamedAB, BC, and CA. Order does not matter, so AB and BA describethe same line segment. The angle between segments AB and B C is labeledABC.

These notes were made using the LaTeX math symbols of AMS TeX.Anyone who has posted hundreds of pages of LaTeX notes has probablydiscovered hundreds of typos and left undiscovered scores of others. I amno exception. If you discover typos please report them to me by email [email protected].

D. J. W.Albuquerque, NMJanuary, 2011


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Contents

1 Two-Body Orbital Mechanics 1

1.1 Keplers Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Newtons Laws . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2.1 An Example . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 History . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3 The Equation of Motion for Two Orbiting Bodies . . . . . . . 91.3.1 Choosing a Goal . . . . . . . . . . . . . . . . . . . . . 91.3.2 Making Things Complicated . . . . . . . . . . . . . . . 101.3.3 Making Things Simple Again . . . . . . . . . . . . . . 10

1.4 Partial Solutions of the Equation of Motion . . . . . . . . . . 141.4.1 Constants of the Motion . . . . . . . . . . . . . . . . . 141.4.2 The Trajectory Equation . . . . . . . . . . . . . . . . . 17

1.5 Conic Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5.1 Polar Equations of Conics . . . . . . . . . . . . . . . . 191.6 Properties of Conic-Section Orbits . . . . . . . . . . . . . . . . 24

1.6.1 Relating the Constants of the Motion to the Geometryof the Orbit . . . . . . . . . . . . . . . . . . . . . . . . 26

1.6.2 Some Important Properties of Individual Conic Orbits 271.6.3 Relationships Among the Conics . . . . . . . . . . . . . 301.6.4 What is B? . . . . . . . . . . . . . . . . . . . . . . . . 321.6.5 The Eccentricity Vector . . . . . . . . . . . . . . . . . 34

1.7 Canonical Units . . . . . . . . . . . . . . . . . . . . . . . . . . 34

2 Orbit Determination from Observations 392.1 Coordinate Systems . . . . . . . . . . . . . . . . . . . . . . . . 39

2.1.1 Heliocentric-Ecliptic Coordinates . . . . . . . . . . . . 402.1.2 Geocentric-Equatorial Coordinates . . . . . . . . . . . 412.1.3 Right Ascension-Declination Coordinates . . . . . . . . 41

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6 CONTENTS

2.1.4 Perifocal Coordinates . . . . . . . . . . . . . . . . . . . 42

2.2 Classical Orbital Elements . . . . . . . . . . . . . . . . . . . . 422.3 Determining the Orbital Elements fromr andv . . . . . . . . 442.3.1 Three Fundamental Vectors . . . . . . . . . . . . . . . 442.3.2 Solving for the Orbital Elements . . . . . . . . . . . . . 45

2.4 Determiningr andv from the Orbital Elements . . . . . . . . 492.5 Coordinate Transformations . . . . . . . . . . . . . . . . . . . 51

2.5.1 Transformations Change the Basis Vectors . . . . . . . 512.5.2 Simple Transformations . . . . . . . . . . . . . . . . . 522.5.3 More Challenging Transformations . . . . . . . . . . . 55

2.6 Mechanics on the Rotating Earth . . . . . . . . . . . . . . . . 632.6.1 An Introduction to Time . . . . . . . . . . . . . . . . . 63

2.6.2 Position, Velocity, and Acceleration . . . . . . . . . . . 672.7 The Ellipsoidal Earth . . . . . . . . . . . . . . . . . . . . . . . 72

2.7.1 The Measurement of Latitude . . . . . . . . . . . . . . 732.7.2 Station Coordinates . . . . . . . . . . . . . . . . . . . . 73

2.8 The Ground Track of a Satellite . . . . . . . . . . . . . . . . . 852.8.1 Launch Site, Launch Azimuth, and Orbital Inclination 85

3 Real Orbits and Orbital Maneuvers 89

3.1 Some Types of Orbits . . . . . . . . . . . . . . . . . . . . . . . 893.1.1 Classification of Earth Orbit by Altitude . . . . . . . . 89

3.1.2 Classification of Orbits by Inclination . . . . . . . . . . 913.1.3 Sun-Synchronous Orbits . . . . . . . . . . . . . . . . . 923.1.4 Special Orbits . . . . . . . . . . . . . . . . . . . . . . . 92

3.2 The Earths Equatorial Bulge . . . . . . . . . . . . . . . . . . 933.3 In-Plane Orbit Changes . . . . . . . . . . . . . . . . . . . . . 94

3.3.1 Launching a Satellite and Adjusting its Orbit . . . . . 943.3.2 Hohmann Transfer . . . . . . . . . . . . . . . . . . . . 963.3.3 General Coplanar Transfer . . . . . . . . . . . . . . . . 100

3.4 Bi-elliptic Transfer . . . . . . . . . . . . . . . . . . . . . . . . 1063.5 Out-Of-Plane Orbit Changes . . . . . . . . . . . . . . . . . . . 111

4 r and v as Functions of Time 1134.1 What we Have Done . . . . . . . . . . . . . . . . . . . . . . . 1134.2 Elliptical Time of Flight as a Function ofE . . . . . . . . . . 114

4.2.1 Two Approaches: Geometric and Analytical . . . . . . 1144.2.2 Why is This Integral so Difficult? . . . . . . . . . . . . 116


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CONTENTS 7

4.2.3 Keplers Geometric Method - Developing Keplers Equa-

tion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1174.2.4 Time of Flight Between Arbitrary Points . . . . . . . . 1214.2.5 Analytical Method . . . . . . . . . . . . . . . . . . . . 123

4.3 Parabolic Time of Flight as a Function ofD . . . . . . . . . . 1254.4 Hyperbolic Time of Flight as a Function ofF . . . . . . . . . 127

4.4.1 Area as a Measure of Angle . . . . . . . . . . . . . . . 1274.4.2 Hyperbolic Time of Flight . . . . . . . . . . . . . . . . 131

5 Orbit Determination from Two Positions and Time 137

6 Ballistic Missile Trajectories 139

6.1 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1396.2 Purpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1406.3 Basics of Ballistic Missiles . . . . . . . . . . . . . . . . . . . . 140

6.3.1 The Non-dimensional Parameter, Q . . . . . . . . . . . 1416.3.2 The Free-Flight Range Equation . . . . . . . . . . . . . 1416.3.3 The Flight-Path Angle equation . . . . . . . . . . . . . 142

7 Lunar Trajectories 145

7.1 Sphere of Influence . . . . . . . . . . . . . . . . . . . . . . . . 1457.2 The Patched Conic Approximation . . . . . . . . . . . . . . . 146

8 Interplanetary Trajectories 147


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8 CONTENTS


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Chapter 1

Two-Body Orbital Mechanics

A story has to start somewhere. Our story starts with Keplers Laws.

1.1 Keplers Laws

Following our text, Fundamentals of Astrodynamics by Bate, Mueller, andWhite, we start with Keplers Laws of Planetary Motion, which are general-izations derived from the planetary position data of Tycho Brahe. Accordingto our text, Kepler published the first two laws in 1609, and the third in 1619.

Keplers Laws

Keplers First Law - The orbit of each planet is an ellipse with the Sun atone focus.

Keplers Second Law - The line joining any planet to the Sun sweeps outequal areas in equal times.

Keplers Third Law- The squares of the periods of any two planets are inthe same proportion as the cubes of their mean distances from the Sun.

These laws explain what the orbits are. Their shapes are ellipses, and thelocal, instantaneous speeds within the ellipses change so that the second andthird laws are true. These laws do not explain why the shapes and speedsof the orbits behave in this way. Our first goal will be to apply the work ofNewton to understand why.

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2 CHAPTER 1. TWO-BODY ORBITAL MECHANICS

1.2 Newtons Laws

Newton did his fundamental work in the 1660s, but did not publish untilthe Principia appeared in 1687. Without the help of Edmund Halley thePrincipia might never have seen the light of day. The Principia containsNewtons three Laws of Motion and the Law of Universal Gravitation (moreon that later). You will find variations in the statements of the Laws ofMotion because the Principia is in Latin, and there is more than one Englishtranslation.

Newtons Laws of Motion

Newtons First Law - A body at rest stays at rest, and a body in motion

stays in uniform, straight-line motion, unless acted upon by a net force.Newtons Second Law - The time rate of change of momentum is propor-tional to the impressed force and is in the same direction as that force, ordpdt

= Ftot.

Newtons Third Law - For every action there is an equal and oppositereaction, or, forces always come in pairs. If body A exerts a force on bodyB, then body B exerts a force on body A that is equal in magnitude andopposite in direction.

The First Law corrects a fallacy in Aristotles physics, that the naturalstate of motion is rest. Newton says that the natural state of motion is

straight-line, uniform motion, that is, motion with constant linear momen-tum. Rest is a special case in which the momentum is zero. It is interestingthat most students naturally believe in Aristotles physics, even if they havenever heard of it. They work hard in Physics 121 to overcome this naturalbelief.

The Second Law is effectively a definition of force from momentum, orfrom mass and acceleration. It explicitly involves the time rate of change, soit assumes that the reader is familiar with calculus. Let us remind ourselvesof the definition of momentum,

p= mv,

so thatdp

dt =

d(mv)

dt =m

dv

dt+ v

dm

dt = Ftot,

where Ftot is the total force.


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1.2. NEWTONS LAWS 3

Force and momentum are vectors. This is not stated explicitly in the

Second Law, I have added that part. Newton does say that the directionof the momentum change and the impressed force are the same, implyinga vector relationship. The term impressed force is a bit vague, and someimpressed forces can be counteracted, at least partially, by friction, whichmay be passive. Some statements of the Second Law use the phrase net forceor unbalanced force, which may impart more clarity. The force that causesa change in momentum is one that is notcounteracted or nullified by otherforces.

The fact that forces are vectors is under appreciated, and its proof is tooeasily ignored. The statement that the time rate of change of momentumequals the total force implies a vector sum of forces.

The fact that F, dpdt , and ma are all vectors means that they can, and,in application, must, be resolved into components. In every application ofNewtons Laws we will require ourselves to choose and explicitly state a co-ordinate system, unless there is a compelling reason not to. We will makefrequent use of Cartesian coordinates and spherical polar coordinates. Oneof our challenges will be to develop the definition of what a coordinate trans-formation is and what it does.

The Third Law turns out to be the one that challenges students the most,even more than the First Law. This is because, while forces do come in pairs,the two forces act on different bodies. This must be fully understood beforethe Second Law is applied. We use free body diagrams as a bookkeepingtechnique to assure that the bodies on which the forces act are properlyassigned.

We need to work at least one example, but first a reminder. Bodiesmay exert forces on themselves, but these forces do not cause a change inmomentum. Consider the classic example of pulling ones self up by onesbootstraps. If you pull up on your bootstraps, your bootstraps must pulldownward on you. By the Third Law these forces must be equal and oppo-site. So far, so good. Assuming that your boots do not rip apart and thatyour boots are are attached to your feet, the forces cancel and no change inmomentum results.

When do forces result in change in momentum? Forces cause a changein momentum when the total force or net force is nonzero, dp

dt= 0 when

Ftot= 0.

Problem 1 The Third Law


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If you cannot exert a net force on yourself how can you stand up from

your chair at the end of class?

1.2.1 An Example

As an example, consider a very simple railroad train, consisting of one engine,one freight car, and one caboose. Let the train operate on a straight, leveltrack. We choose a two-dimensional coordinate system withxalong the trackandy upward, perpendicular to the track. By convention we must choose aright-handed coordinate system, so letx be to the right. Let us assume thatthe train cars individually are of constant mass, so we know from the SecondLaw that

dp

dt = F =ma. (1.1)

This is a vector equation, so a similar equation must apply in each coordinatedirection,

Fx= max, (1.2)

andFy =may. (1.3)

There is no motion in the zdirection, so we may choose to ignore the ap-plication of the Second Law there. If we do apply the Second law to the zdirection the result is trivial because Fz = 0 and az = 0. Even with thissimplification we will see that a thorough analysis of the motion of the trainbecomes complex very rapidly.

By convention, we draw a free-body diagram for each of the three railroadcars. The argument above shows that we need to be concerned with thexand y directions only. Consider the engine first. At least four forces acton the engine. Three are obvious - the engines weight due to gravity, theupward force of the track on the engine (in the y direction), and the forwardor backward force of friction of the track on the engine (in either the plusor minusx direction). The engine and track are in contact where the wheels

touch the track, so that is where the friction and the upward force act. Thefriction may be forward or backward according to what the engineer is doingwith the drive motor and the brakes. There must be some friction or thetrain could not move. Judicious application of the power of the drive motormust move the train forward, similar use of the brakes must slow it down.


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1.2. NEWTONS LAWS 5

Lets assume that the engineer is controlling the power so that the train is

moving forward.The fourth force is not so obvious. The engine exerts a force on thefreight car, thus pulling the freight car forward, so the freight car must exerta backward force on the engine according to the Third Law. This is thefourth force. Thus, the free-body diagram of the engine shows four forces:the weight downward, the track normal force upward, a frictional force fromthe track that is forward, and a backward force exerted by the freight car onthe engine.

Next, consider the freight car. It, too, has four forces in its free-bodydiagram. There is still a weight downward and a track force upward. Theengine pulls the freight car forward. The freight car is attached to the ca-

boose, and the freight car pulls the caboose forward, so by the Third Lawthe caboose must exert a backward force on the freight car. These are thefour forces. We could include a fifth force, the friction of the track on thecar, but we choose to ignore it.

Finally, consider the caboose. It has only three forces in its free-bodydiagram, because it is not attached to anything behind it. There is still aweight, still a track force upward, and the force exerted by the freight car onthe caboose. Again, we choose to ignore friction.

We have all watched or ridden in trains that operate on a track that isessentially straight and level. Such a train never sinks into the Earth or fliesupward from the track, so we infer that a

yis always zero, or very nearly so.

By application of the Second Law, this means that the sum of the forces inthey direction must also be zero, so the weight of the engine and the upwardforce on the engine exerted by the track must be balanced - they must beequal and opposite, so their sum is zero. This conclusion applies to all of thecars in the train. It means that all of the interesting things that happen arein the x direction.

Apply the Second Law to the engine using its free-body diagram as aguide. In the positive x direction we have the force of friction on the engine,fe. In the negative x direction we have the force of the freight car on theengine,Ffce. The Second Law gives

fe Ffce = meae, (1.4)where me is the mass of the engine and ae is its acceleration. An applicationof the Second Law such as this gives the equation of motion, in this case theequation of motion for the engine.


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Similarly, apply the Second Law to the freight car to get its equation of

motion, Fefc Fcfc = mfcafc, (1.5)whereFefcis the force of the engine on the freight car, Fcfc is the force of thecaboose on the freight car, mfc is the mass of the freight car, and afc is itsacceleration. Then apply the Second Law to the caboose to get its equationof motion,

Ffcc = mcac, (1.6)

where Ffcc is the force of the freight car on the caboose, mc is the mass ofthe caboose, and ac is its acceleration.

Taken together these three equations seem daunting; they certainly do

not invite solution. They will, once they are simplified. First, note that thecars of the train are rigidly tied together by the couplers, so they must allhave the same acceleration,

ae=afc = ac= a. (1.7)

Next, notice that by the Third Law the force exerted by the freight car onthe engine must be equal and opposite to the force exerted by the engineon the freight car. The directions have been accounted for in the free-bodydiagrams, so we need only concern ourselves with the magnitudes of theforces. We write Ffce = Fefc = F1. A similar application of the Third Law

shows that the force of the caboose on the freight car must be equal andopposite to the force of the freight car on the caboose. Again, direction hasbeen accounted for in the free-body diagrams, so the magnitudes of the forcesbecome Fcfc = Ffcc = F2.

With these simplifications we can rewrite the equations of motion. Theequation for the engine becomes

fe F1 = mea, (1.8)

that of the freight car becomes

F1 F2 = mfca, (1.9)and that of the caboose becomes

F2= mca. (1.10)


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1.2. NEWTONS LAWS 7

So far we are left with three equations in the four unknowns a, F1, F2, and

ff. If we can eliminate one of the unknowns then we can solve for the motionof the train. This is good. We know that real trains move, and we are gettingclose to understanding why.

There are two obvious ways to proceed. One is to measure the accelerationof a real train and calculate the forces involved. The other is to measure thefriction and calculate the acceleration and the forces among the cars. Wechoose to measure the friction.

In many situations the friction is well described by ff = N, where is the coefficient of friction and N is the normal force. The coefficient offriction is taken to be a constant of order unity that depends on the natureof the materials that are in contact. The normal force is simply the upward

force exerted by the track. In the case of the engine we have shown that itis merely the engines weight, or meg, where g is the acceleration of gravity,so ff =meg. We can plug this into the equation of motion for the engine,getting

meg F1 = mea. (1.11)This presents a strategy for solving for the motion of the train: solve thisequation forF1, plug that into the equation of motion for the freight car andsolve forF2. Plug that into the equation of motion for the caboose and solvefor a. The value ofa can then be used to fine F1 and F2. Proceeding withthe plan,

F1= meg mea= me(g a). (1.12)Plugging into the equation of motion for the freight car,

meg mea F2 = mfca, (1.13)

and solving for F2,

F2= meg meamfca= meg (me+mfc)a. (1.14)

Plugging into the equation of motion for the caboose,

meg (me+mfc)a= mca, (1.15)

or

meg= (me+mfc+mc)a= mtraina. (1.16)


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This says that the friction between the engine and the track must be large

enough to accelerate the entire train! Amazing! Solving for the acceleration,

a= meg

mtrain, (1.17)

which says that the acceleration is the friction force divided by the total massof the train. This makes sense. Note that if there is no freight car or caboose,so their masses are each zero, then

a=meg

me=g, (1.18)

which may make sense based on your earlier studies.

Notice how complicated and subtle this analysis is. Lets take a look atwhat we have accomplished. We have calculated the motion of a train fromfirst principles using Newtons Laws. At the beginning of the seventeenthcentury nobody was able to do this. By the end of the seventeenth centurythe most skillful mathematicians and natural philosophers were able to dothis. Trains were unknown at the time, so they would have analyzed a horsepulling a wagon, but the analysis would have been the same.

1.2.2 History

What historical change allowed this analytical approach to happen? Mathwas applied to the description of the physical world. The crucial step beforethe work of Newton was made by Galileo. He had mentors and spoke withthem about what they were doing, but Galileo showed, by laboratory mea-surement, that an object undergoing constant acceleration has equal velocitychanges in equal amounts of time. This is the basis for the definition ofconstant acceleration familiar to us, a = dvdt =const., and led to the develop-ment of calculus by Newton and Leibniz. Previous to Galileos measurementsand interpretation the weight of opinion was that accelerating objects gainedequal amounts of velocity in equal distances, a = dvdx , which we know to beincorrect. This opinion was based on pure thought, not on laboratory mea-

surement. Galileos discovery is described in the Dialogues Concerning TwoNew Sciencesin the section called Day Three.

That was not Galileos only accomplishment in the Dialogues. In thesection called Day Two he applied algebra to the description of the bendingand failure of beams. Many modern critics have pointed out that his analysis


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1.3. THE EQUATION OF MOTION FOR TWO ORBITING BODIES 9

is only partially correct because he did not draw a free-body diagram of

the beam and was unclear about the balance of compression and tension atany cross section of a stationary beam. His major accomplishment was theperformance of measured experiments and the application of algebra to theirresults.

1.3 The Equation of Motion for Two Orbiting

Bodies

Newtons formulation of gravitation also appeared in the Principia, and we

know it as Newtons Law of Universal Gravitation. For two bodies of massesM and m it is written as

Fg = GMmr2

r

r = GMm

r3 r, (1.19)

where Fg is the force of gravity that one of the bodies exerts on the other, Gis a constant of nature,ris the vector from the body causing the force to thebody experiencing it, and r is the magnitude or the vector r. The negativesign makes the force attractive. We credit Henry Cavendish, 1731-1810, withmeasuringG, the gravitational constant. He would have claimed that he didsomething very different - that he determined the mass of the Earth, andfrom this its density. This result, in itself, is profound, for Cavendish showedthat the overall density of the Earth is very similar to that of iron or nickel,suggesting that they are the main constituents of the solid Earth, and, byimplication, that the oceans have very limited depth.

Owing to the definition ofr, the force exerted by M on m is equal andopposite to the force exerted bymonM, as required by the Third Law. Thiswill be important in deriving a simple orbit equation.

1.3.1 Choosing a Goal

We would like to know the position as a function of time of orbiting objects,and we would like to derive them from first principles, such as our knowledgeof Newtons Laws and Universal Gravitation. This is a challenging goal.


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1.3.2 Making Things Complicated

In general, Fg is only one of several forces that contribute to the total forceacting on an orbiting body. Other forces may include gravitation by otherbodies, radiation pressure on the cross-sectional area of the orbiting body,drag owing to the low density of the residual atmosphere, and any other forcethat can be documented or hypothesized. In general

dp

dt = Fnet, (1.20)

where Fnet is the sum of all acting forces.

1.3.3 Making Things Simple Again

Not wishing to deal with all of this potential complication, lets make somesimplifying assumptions and examine the resulting special case.

Assumptions:

Assumption 1. We assume that we have an inertial coordinate system(X, Y , Z ) that is adequate for describing the system under study.

An inertial coordinate system is one that is not accelerating. The Earthis spinning on its axis and orbiting the Sun, and the Sun is orbiting the

Galaxy, so our location on the surface of the Earth is not an inertial frame.The accelerations must be very small, because we treat the surface of theEarth as an inertial frame all the time.

Problem 2 How nearly inertial is the Earths reference frame?Calculate the accelerations of the Earth due to the Earths rotation, the

Earths orbit about the Sun, and the Suns orbit about the Milky Way.Assume that all of these motions are circular, and that a = v2/r. Work thisproblem in mks units and compare the accelerations with that of gravity atthe Earths surface. You will need the radius of the Earth, the radius ofthe Earths orbit, and the radius of the Suns orbit. The first two should

be familiar, assume that the third is 8 kiloparsecs and convert to meters.The velocities can be found from the radii and orbital periods. Again, thefirst two should be familiar, assume that the third is 250,000,000 years. Inconverting to seconds, show that the number of seconds in a year is verynearly 107.


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Following Figure 1.2-1 in the text, we number the masses under study in

this system from 1 to n, so their masses arem1, m2, ...mi, mj, mk,...mn (1.21)

and their position vectors are

r1, r2, ...ri, rj, rk, ...rn. (1.22)

Problem 3 You should convince yourself that the vector from mn to mi,calledrni in the text, is

rni= ri rn. (1.23)

You should do this in three ways: by settingri = 0, by settingrn = 0, andby making a sketch of the vectors in (X, Y , Z ) space.

I am somewhat displeased with the notation used in the text, because thesubscriptn carries a special meaning attached to the last named mass. I pre-fer replacingn withj , which has no special meaning, to make the statementmore general. Thus, I would write

rji = ri rj . (1.24)

Continuing with the general notation, this means that the force exerted on

massmi by mass mj isFgji= Gmimj

r3jirji. (1.25)

Here I have changed the notation slightly. The text would call this forceFgj , I have added the additional subscript i for clarity. The sum of all suchgravitational forces acting on massmi by all possible masses mj is then

Fgi = Gmim1r31i

r1iGm1m2r32i

r2i ... Gminnr3ni

rni = Gmin

j=1

mjr3ji

rji , (1.26)

which partially returns us to the notation of the book. Here we face a smallproblem: what do we do whenj = i? Can a body exert a force on itself? If itcan, this seems to imply that the body can be subdivided into two interactingparts, in which case those parts must exert equal and opposite forces by theThird Law. If so, the forces cancel in the sum. Thus, we take as given that


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a body cannot exert a net nonzero gravitational force on itself until we can

be convinced otherwise. The total gravitational force on mass i becomes

Fgi = Gmin

j=i

mjr3ji

rji. (1.27)

We must allow for the possibility that gravity is not the only force, sothat Ftot = Fgrav+ Fother, where the other force is yet to be specified. Thistotal force then equals the time rate of change of momentum.

Assumption 2. We assume that only two bodies are present in this spacethat is otherwise empty, or so nearly empty that emptiness may be assumed.Thus,i= 1, 2.

Assumption 3. We assume that the bodies are spherically symmetric sothat their gravitation is mathematically identical to that of point masseslocated at their centers. This assumption makes gravitational forces centralforces. To say this another way, the gravitational forces are then antiparallelto the radius vectors, so any cross product, like that for torque,=ri Fg,must be zero. This means that gravity cannot cause torques. We will have torelax this assumption in Chapter 3 when we consider the Earths equatorialbulge, which has a non-spherical mass distribution.

Assumption 4. We assume that the bodies have constant masses, so

dpidt

=midvidt

=mi dri

dt =miri. (1.28)

This turns out to be a greater help to us than might originally be imag-ined. If the masses are constant then the number of time-dependent variablesthat we wish to solve for is reduced from eight to six, that is, from one massand three positions for each object to just the three positions for each object.

Letr be the vector from M tom. Following the notation in the text wewill let body 1 have mass M, so m1 = M, and let body 2 have mass m, som2= m. Let the position ofMber1 and that ofm be r2. The position ofmrelative toM is thenr21=r1r2 = r, and the position ofMrelative tomisr12 = r2

r1 =

r. These vectors can be used to calculate the gravitational

forces.

The gravitational force exerted by M on m is

mr2 = GMmr2

r

r, (1.29)


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and the gravitational force exerted by m on M is

Mr1 =GMm

r2r

r. (1.30)

The corresponding accelerations are

r2 = GMr3

r, (1.31)

and

r1=Gm

r3 r. (1.32)

Subtracting the second acceleration from the first gives

r2 r1= r= G(M+ m)r3

r. (1.33)

This equation involves only the relative position,r, and converts the problemof gravitational motion into an equivalent one-body problem, with a bodywhose mass is the sum of the two masses. We now have only three unknowns,the three vector components ofr, which is a great simplification.

It is customary to define = G(M+ m), and write

r+ r3

r= 0. (1.34)

This is the equation that will form the basis of our work for the remainderof the semester. In nearly all the cases that we will study Mis much greaterthanm, so GM.Assumption 5. We assume that the bodies have distinctly different masses,with the more massive body having mass Mand the less massive body havingmass m, such that M m. This assumption sets us up for cases likeplanetary motion, in which M is the mass of the Sun and m is the mass of

a planet, or satellite motion, in which Mis the mass of the Earth and m isthe mass of the satellite. In either case M m is a very good assumption.If this assumption breaks down then we will have to modify the subsequentequation. The assumption does not apply to binary star systems in whichthe two stars have similar masses.


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1.4 Partial Solutions of the Equation of Mo-

tion

1.4.1 Constants of the Motion

Specific Mechanical Energy

Our goal is a solution to the equation of motion, that is, the functional formforr(t). This is a lofty goal that will require a lot of work. It turns out to berelatively easy to learn a lot about r(t) without actually finding the solution,and that is how we will start.

To someone experienced in solving the equation of motion it is natural

to multiply both sides by rand integrate. This may look like a trick at first,but it is not, for many equations of motion can be integrated in this way.

Experience shows that it is an obvious thing to do with an obvious result.Dot multiply on the left by r to get

r r+r r3

r= 0. (1.35)

We know that r =v, and we can show that a a=aa. Rather than assignthis as homework, here is the proof. We know

a

a= a2, (1.36)

sod

dt(a a) = d

dta2, (1.37)

and2a a= 2aa, (1.38)

soa a= aa. (1.39)

Now our equation becomes

v v+ r3

r r= 0, (1.40)

giving

vv+

r3rr= 0. (1.41)


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1.4. PARTIAL SOLUTIONS OF THE EQUATION OF MOTION 15

This can be integrated by inspection, because

d

dt

v2

2 =vv, (1.42)

andd

dt

r

=

r2r. (1.43)

Try these for yourself. This gives

d

dt

v2

2

r

= 0. (1.44)

This result can be made even more general by noticing that

d

dt

v2

2 +c

r

= 0 (1.45)

where c is an arbitrary constant. Since the net time derivative is zero wemay take the function in parentheses to be a constant of time and write

E= v2

2 +

c

r

(1.46)

where Eis a constant. The first term in Eis the kinetic energy per unit mass,or specific kinetic energy. The second term is the gravitational potentialenergy per unit mass (or the specific potential) plus an arbitrary constant.If we choose the constant to be zero then the specific potential goes to zerofrom below as r becomes large. Choosing c = 0 leaves us with the specificmechanical energy,

E= v2

2

r, (1.47)

which is a constant of the motion. Another way of saying this is that thespecific mechanical energy is conserved.

Specific Angular Momentum

Experience shows that the dot multiplication above is an obvious thing to doand reliably leads to the conservation of energy. It is less obvious to try cross


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multiplication byr, but it can do no harm, and will eliminate one term from

the equation of motion because any vector crossed with itself gives zero. Try

r r+ r r3

r= 0. (1.48)

The second term must be zero, leaving

r r= 0. (1.49)

This, too, can be integrated by inspection by trying

d

dtr

r =r r+ r

r. (1.50)

Any vector crossed with itself must again give zero, leaving the other term,so

d

dt

r r

=

d

dt

r v

= 0. (1.51)

Thus, we have discovered another constant of the motion, rv. This is justthe specific angular momentum, and we write

h= r v. (1.52)

This result turns out to be more subtle than the previous one, for h mustbe constant in both magnitude and direction, forcing the vectors r and vto define a plane, since h must be perpendicular to both r and v by thedefinition of the cross product. Thus, the orbital motion of the two bodies isconfined to a plane when specific angular momentum is conserved.

Writing out the magnitude of the cross product gives

h= rv sin =rv cos , (1.53)

where is the angle between r andv when they are drawn tail to tail, and is the complement of. Please refer to Figure 1.4-1 in the text. is the

angle between the local vertical and v, and is called the zenith angle. Thus, is the angle between the horizontal andv, and is called the flight-path angle.Of these two angles is usually the more easily observable, and will appearin future chapters, so it it convenient to introduce it here. The sign of isthe same as the sign ofr v.


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1.4. PARTIAL SOLUTIONS OF THE EQUATION OF MOTION 17

1.4.2 The Trajectory Equation

The next manipulation of the equation of motion is brilliant, very subtle,and not at all obvious. It was first performed by Newton, and I wonder howhe did it so long ago, before our current vector notation had been developed.Write the equation of motion as

r= r3

r, (1.54)

and cross multiply by h from the right, while at the same time changingorder of operation and sign on the right-hand side of the equation, to get

r

h=

r3 h r. (1.55)

Problem 4 The time derivative of the cross productCompare the left-hand side of the equation with

d

dt

r h (1.56)

to show that they are equal.

The right-hand side can also be expressed as a time derivative. Expandhto get

r3h r =

r3

r v r= r3

v

r r rr v

=

rv r

r2r. (1.57)

The next-to-last step makes use of a vector identity that appears in AppendixC of the text. There are two closely-related identities, the first of which Iknow as the B AC CAB rule, the name providing an aid to memory:

A ( B C) = B( A C) C( A B), (1.58)

and( A B) C= B( A C) A( B C). (1.59)

We note that

d

dt

r

r

=

rv r

r2r. (1.60)


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Thus, we can writed

dt

r h =ddtrr. (1.61)Integrating both sides gives

r h= rr

+ B, (1.62)

where Bis a vector constant of integration, and turns out to be an additionalconstant of the motion. Dot multiplying byr on the left gives

r r h=r rr

+ r B. (1.63)

There is a vector identity

a b c= ab c, (1.64)and we already knowa a= a2, so

r r h= r+ r B, (1.65)giving the scalar equation

h2 =r+rB cos , (1.66)

where is the angle between B andr . Solving for r gives

r= h2/

1 + (B/)cos . (1.67)

Very nice, but what has this accomplished?

1.5 Conic Sections

Conic sections are formed by the intersection of a cone and a plane, as in

Figure 1.5-2 in the text. A true mathematical cone looks like two ice creamcones set parallel with points touching. We will see that conic sections includecircles, ellipses, parabolas, hyperbolas, lines, and a point. Conic sections wereknown and studied by the Greeks. Knowledge of conic sections would havebeen common among educated people in the time of Kepler and Newton.


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1.5. CONIC SECTIONS 19

The following derivations draw heavily on material from Wolfram Math-

World, cited as:Weisstein, Eric W. Ellipse. From MathWorld A Wolfram Web Resource.http://mathworld.wolfram.com/Ellipse.html

Weisstein, Eric W. Hyperbola. From MathWorld A Wolfram Web Re-source. http://mathworld.wolfram.c/Hyperbola.html

and

Weisstein, Eric W. Conic Section Directrix. From MathWorld A WolframWeb Resource. http://mathworld.wolfram.com/ConicSectionDirectrix.html

Additional valuable material on ellipses may be found at

http://www.oc.nps.navy.mil/garfield/ellipse app2.pdf

1.5.1 Polar Equations of Conics

The equation of an ellipse with the coordinate origin located at its center is

x2

a2+

y2

b2 = 1, (1.68)

wherea is the semi-major axis and b is the semi-minor axis. This is probablyfamiliar from your earlier studies. This is not the only way to describe anellipse, and, in fact, we seek a description in polar coordinates with the originat a focus of the ellipse. The motivation for this is a desired comparison withthe trajectory equation. Remember Keplers First Law? The orbit of eachplanet is an ellipse with the Sun at one focus. If the Sun is the center of theSolar System then its center is the logical place to put the origin of a set ofpolar coordinates, so we wish to move the coordinate origin to a focus. Fromthe diagram we see that

x= c +r cos , y= r sin . (1.69)

Making the obvious substitutions and multiplying both sides by a2b2 gives

b2c2 + 2b2cr cos +b2r2 cos2 +a2r2 sin2 = a2b2. (1.70)

We are going to manipulate this equation and compare the result with thetrajectory equation. We will see that the two equations are the same. The


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web sites listed above show that there are several parameters other thana and

b that are used to describe ellipses. Two common ones are the eccentricity,e, and the half-focal separation, c, which is the distance from the center ofthe ellipse to either focus. Mathematically,

e=

1 b

2

a2, (1.71)

andc= ae. (1.72)

We use these expressions to eliminate b andcby replacing them with termsina and e, using

b2

=a2

(1 e2) (1.73)and the trigonometric identity

sin2 = 1 cos2 (1.74)

to get

a2(1 e2)a2e2 +a2r2 + 2a2(1 e2)aer cos +a2(1 e2)r2 cos2 a2r2 cos2

=a2a2(1

e2). (1.75)

Dividing through by a2 is an obvious step. Regrouping gives

r2 +a2(1 e2)(e2 1) + 2a(1 e2)er cos e2r2 cos2 = 0. (1.76)

Next multiply through by1 and isolate the r2 term to get

r2 =a2(1 e2)2 2a(1 e2)er cos +e2r2 cos2 . (1.77)

Taking the square root gives

r=er cos a(1 e

2), (1.78)where the sign is to be determined. The radius and a must be positive, andemay be positive or zero. Taking e = 0 gives

r= (a), (1.79)


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so we must choose the leading negative sign, giving

r= a(1 e2) er cos . (1.80)

Grouping the terms inr gives

r

1 + e cos

=a(1 e2), (1.81)

and solving for r gives

r= a(1 e2)1 + e cos

. (1.82)

This equation is commonly written as

r= p1 + e cos

(1.83)

where p is called the semi-latus rectum, or parameter, and e is the eccentric-ity. Thus,

p= a(1 e2). (1.84)Compare this result with the trajectory equation

r= h2/

1 + (B/)cos . (1.85)

We identify

p=h2

, e=

B

, =. (1.86)

Thus, we have shown that ellipses match the mathematical form of gravita-tional orbits.

This is not the only possibility: in general gravitational orbits are conicsections, which include circles, ellipses, parabolas, and hyperbolas. This ismost easily discovered by examining the directrix definition of conic sections.The conic sections may be defined using a line, called a directrix, and apoint, called a focus. The conic section itself is the locus of points whose

distance from the focus is proportional to the horizontal distance from thedirectrix,given a vertical directrix. (The quote is from Weissteins articleConic Section Directrix, reference at the top of this section. The italicemphasis is mine.) If the two distances are equal then the conic section is aparabola. If the distance to the focus is smaller than the horizontal distance


30/155


to the directrix then the conic is an ellipse. If the distance to the focus is

larger than the horizontal distance to the directrix then the conic section isa hyperbola.It is tempting to geuss that when the distance to the focus equals the

distance to the directrix the result is a parabola. Lets find out. Consider afocus at position (a, 0) and a directrix atx = a, and a general point (x, y).The horizontal distance to the directrix is x + aand the distance to the focusis

(x a)2 +y2. Equate and square both sides to getx2 + 2ax+a2 =x2 2ax+a2 +y2. (1.87)

Subtractingx2 and a2 from both sides leaves

y2 = 4ax, (1.88)

which is the familiar equation of a parabola.Switching to polar coordinates with origin at the focus gives

x= a r cos , y= r sin , (1.89)so the equation of the parabola becomes

r2 sin2 = 4a2 4ar cos . (1.90)Substituting 1

cos2 for the sin2 term and reorganizing gives

r2 = (2a r cos )2. (1.91)Taking the square root gives

r= (2a r cos ). (1.92)We again require that r be positive, and examine the result when = 0 andr= a, giving

a= (2a a), (1.93)where the choice of the positive sign is clear. Then

r= 2a r cos , (1.94)and solving for r gives

r= 2a

1 + cos , (1.95)


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which matches the trajectory equation, with p = 2a and e = 1. Thus,

parabolas also match the trajectory equation.Rather than show that a hyperbola also matches the trajectory equationwe choose to work the general result for any conic. Start with Cartesiancoordinates (x, y) with an originO. Place a focus at (c, 0) and a verticaldirectrix at (c+d, 0). This can describe a general case if we allow d to bepositive or negative. Choose the origin of a second coordinate system at thefocus. Label the coordinates relative to this origin (x, y) and call the originO. Relative to this origin the focus is at (0, 0), the directrix is at x = d,and an arbitrary point has coordinates (x, y). Clearly, the distance of thisarbitrary point from the origin is

x2 +y2 and its horizontal distance from

the directrix is (d

x). We seek the equation in (x, y) coordinates that

defines figures that have a constant ratio of proportionality, e, between thesedistances, so, working with the squares to get positive-definite quantities

x2 +y2 =e2(d x)2 =e2(x2 2xd+d2), (1.96)or,

y2 = (e2 1)x2 2e2dx+e2d2. (1.97)We may also define polar coordinates (r, ) with respect to originO suchthat

x= rcos, y= r sin , sin2 = 1 cos2 , (1.98)

givingr2(1 cos2) = (e2 1)r2 cos2 2e2dr cos +e2d2. (1.99)

Recognize that both sides of the above equation have a termr2 cos2 thatcancels, and reorganize to get

r2 =e2(r2 cos2 2rd cos +d2) =e2(r cos d)2. (1.100)Taking the square root gives

r= e(r cos d). (1.101)

When = 90o we obtainr= e(d), (1.102)

which leads us to choose the leading negative sign, leaving

r= e(d r cos ). (1.103)


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Solving for r gives the desired result,

r= ed

1 + e cos . (1.104)

Problem 5 The polar form of the hyperbolaStarting with the equation for a hyperbola with its symmetry point at

the origin, its foci at (c, 0) and (c, 0), and its vertices at (a, 0) and (a, 0),x2

a2 y

2

b2 = 1, (1.105)

put the equation in the form of the trajectory equation with the origin at the

focus (c, 0). You will need c= ea and e =

1 + b2

a2 . Notice the sign changein the formula for e compared with that for the ellipse.

The conic sections are distinguished by their eccentricities. The circleand parabola are special cases, with e = 0 and e= 1, respectively. Ellipseshave 0< e 1. The conic sections are the onlypossible paths for two-body orbits.

1.6 Properties of Conic-Section Orbits

As mentioned above, conic sections are the only possible orbits for a two-body system. In real systems, e.g., the Solar System, deviations from conicsections are caused by the presence of additional bodies.

We can now understand Keplers First Law: The orbit of each planet isan ellipse with the Sun at one focus. This is a consequence of gravitationallybound, enduring, two-body orbits, and the fact that the Sun is much moremassive than any of the planets. Parabolic and hyperbolic orbits are allowed,but they do not endure because those figures extend to infinity. Actually,the center of mass of the two bodies is at the focus, but the Sun is so massivethat the center of mass is very close to the center of the Sun. We have seen

that the elliptical orbit takes place in a plane that is fixed in space, andthatr andv change in such a way that E,h, and B remain constants of themotion.

A general conic section has two foci. The circle and parabola are specialcases - the two foci are coincident in the circle and one focus is at infinity for


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1.6. PROPERTIES OF CONIC-SECTION ORBITS 25

the parabola. The empty focus appears to have no particular importance in

orbital mechanics. Similarly, the directrix has no particular importance. Theheight of the orbit at the location of the focus is the parameter or semi-latusrectum, p. The length of the orbit along the line defined by the foci is calledthe major axis and has extent 2a, giving the semi-major axis length a. In acircle 2a is the diameter and a is the radius. For the parabola 2ais infiniteand for the hyperbola it is taken as negative. The distance between the fociis 2c. For the circlec is zero, for the ellipse it is positive, for the parabolainfinite, and for the hyperbola negative. Please refer to figure 1.5-3 in thetext. For all conics except the parabola

p= a(1

e2) (1.106)

and

e= c

a. (1.107)

Note that for a parabolae = 1 but p = 0.The extreme points of the orbit are known as turning points or apses, sin-

gular apse, from the Greek via Latin for arch. The point on the orbit nearestthe occupied focus is called the periapsis and the point farthest from the oc-cupied focus is called the apoapsis. The names can change according to the

gravitating body at the focus: perigee and apogee for the Earth, perihelionand aphelion (pronounced afelion) for the Sun, periselenium and aposeleniumfor the Moon (dont get caught off guard by the terms of mixed origin, per-ilune and apolune), and perigalacticon and apogalacticon for galaxies. Noticethat these terms are not uniquely defined for a circle, and that the apoapsishas no definite meaning for a parabola or hyperbola.

We can calculate the distance to the periapsis and apoapsis by setting= 0o and = 180o in the trajectory equation, getting

rmin = rperi = rp = p

1 + e

=a1 e2

1 + e

=a(1

e), (1.108)

and

rmax= rapo = ra= p

1 e =a1 e21 e =a(1 + e). (1.109)


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1.6.1 Relating the Constants of the Motion to the Ge-

ometry of the OrbitThe constants of motion areE, h, and B. The geometry of the orbit isdescribed byp,a,e, and. The quantitiesp,a, ande are not independent -any two can be used to find the third because p = a(1 e2). Comparing thetrajectory equation with the general polar equation for conic section orbitsshows thatp= h2/,e = B/, and that the reference direction for measuring

is provided by the direction of B.

h Determines p - Newtons Cannon

Of the constants of motion, the quantityh alone determinesp. This is obviousfrom the comparison above, which givesp = h2/. This is also demonstratedby a thought experiment proposed by Newton, and illustrated in Figure 1.6-1 in the text. Imagine a cannon on a mountain top. The height of themountain determines r at the moment that teh shot is fired. The cannoncan be fired successively with greater and greater charges of powder. Assumea limitless supply of cannonballs and powder, and ignore air resistance. Aimthe cannon along the local horizontal, so that the flight-path angle, , is zero.Shots using increasing amounts of powder will propel the cannonball in everbroader arcs. More powder increasesv , which increases h, which increasesp.

E Determines aOf the constants of the motion, the quantityEalone determines a. We canshow this by considering periapsis and apoapsis of conic-section orbits. Atthese places the velocity is tangent to the orbit and perpendicular to theradius. Another way of saying this is that the flight-path angle,, is zero.Then

h= rpvp =rava, (1.110)

and

E= v2

2

r =

h2

2r2p

rp, (1.111)

butrp = a(1 e) (1.112)

andh2 =p= a(1 e2) (1.113)


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so

E= a(1 e2)2a2(1 e)2

a(1 e)=

2a

(1 + e)(1 e)

a

1(1 e)

=

2a

1

(1 e) (1 + e 2) =

2a

(e 1)(1 e) =

2a. (1.114)

Thus,E determines a.

p and a, orE and h, Determine eWe already know

p= a(1

e2). (1.115)

Solving fore givese=

1p/a, (1.116)

butp= h2/, a= /2E, (1.117)

which can be used to get the desired result

e=

1 +

2Eh22

. (1.118)

1.6.2 Some Important Properties of Individual ConicOrbits

Elliptical Orbits and Keplers Laws

We know h = r v. The magnitude ofh is the magnitude ofr times thetangential component ofv, which is r. This is illustrated in Figures 1.7-2and 1.7-3 in the text. Thus

h= r2d

dt (1.119)

and

dt= r

2

hd. (1.120)

But the area swept out by the radius vector is

dA=1

2r2d, (1.121)


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so

d=

2

r2dA, (1.122)

and

dt=r2

h

2

r2dA=

2

hdA. (1.123)

This is Keplers Second Law - the line joining any planet to the Sun sweepsout equal areas in equal times. It is a consequence of conservation of angularmomentum.

If we integrate both sides of the above equation over one complete ellip-tical orbit we getA = aband

Tp = 2abh , (1.124)

where Tp is the orbits period, but

b=

a2 c2 =

a2 a2e2 =

a2(1 e2) = ap. (1.125)

We already know that h =

p, so

Tp = 2a

app

= 2

a3/2, (1.126)

orT2p =

42

a3. (1.127)

This is Keplers Third Law - the squares of the periods of any two planets arein the same proportion as the cubes of their mean distances from the Sun.We have shown that Keplers Laws are a consequence of the conic nature ofgravitational orbits. This is quite an accomplishment!

Problem 6 The mass of the SunReminding ourselves that when the mass of the gravitating body is much

greater than the mass of the orbiting body, as in the case of planetary motionin the Solar System, = GM. In such a case Keplers Third Law can besolved for M, giving a method to calculate the mass of the Sun. Use theknown values ofT and a for the Earth to calculate M.


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Circular Orbits

Circles are a special case of ellipses in which a = b = r. We define thecircular speed, vcs, the circular radius, rcs, and the orbital period at thecircular speed, Tcs. Then we can find the circular speed in two ways. First,

Tcs = 2

r3/2cs =

2rcsvcs

, (1.128)

from the circumference divided by the speed, or

E= 2rcs

=v2cs

2

rcs, (1.129)

from the energy equation. Either can be solved to give

vcs=

rcs. (1.130)

Note the importance of Problem 1.17 in the text and Newtons cannon tothe comparison of circular and elliptical orbits.

Parabolic Orbits

Parabolic orbits have two useful properties. First, a parabolas radius atperiapsis is especially simple, as shown in Figure 1.9-1 of the text. Because

e= 1 the orbit in focal polar coordinates is

r= p

1 + cos . (1.131)

Periapsis occurs when = 0o, where cos = 1, so

r= rperi = rp =p

2. (1.132)

The parabolic orbit has zero specific mechanical energy, making it easy tocalculate the escape speed,

v2esc

2

r

= 0, (1.133)

so

vesc=

2

r . (1.134)

At escape vesc goes to zero and r goes to infinity, which agrees withE= 0.


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Hyperbolic Orbits

An important property of hyperbolic orbits comes from the fact that the in-coming and outgoing parts of the orbit have a pair of straight-line asymptotesthat are separated by the turning angle,, which can be calculated becausesin

2= 1e . This is shown in Figure 1.10-1 of the text.

The hyperbolic orbit has positive specific energy, so at any distance rfrom the force center the object in hyperbolic orbit is moving faster than anobject in a parabolic orbit. This gives rise to the hyperbolic excess speed.Imagine putting an object into hyperbolic orbit by burning a rocket engine.We compare the burnout point with infinity using the energy equation,

E=v2bo

2

rbo =v2

2

r , (1.135)

which gives

v2 = v2bo

2

rbo=v2bo v2esc. (1.136)

This is the hyperbolic excess speed.

1.6.3 Relationships Among the Conics

We really have two general types of orbits - elliptical and hyperbolic - and

two special cases - circular and parabolic. A parabolic orbit is a specialintermediate case between elliptical and hyperbolic. A circular orbit is thelimiting case of an ellipse whenbapproachesa. Lets compare the two generalcases, ellipses and hyperbolas.

An ellipse can be drawn using two foci separated by a distance of 2 cconnected by a piece of string of length 2a with a > c. Put a pencil in thestring, keep both segments of string tight, and move the pencil to draw anellipse. When the pencil is on the line bisecting the foci and perpendicularto the line joining them then it defines two right triangles, each of base c,height b, and hypotenuse a, so

a2 =b2 +c2. (1.137)

The arms of hyperbolas are limited by asymptotes that can be used to de-fine right triangles. The foci of the two branches of a hyperbola are separatedby a distance of 2c, as for the ellipse. The apexes are the two branches are


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separated by 2a, as are the apses of the ellipse, but the foci of the hyperbola

are both outside the apexes, so c > a, and the situation is reversed relativeto that of the ellipse. An arc of a circle of radius c with its center at theintersection of the asymptotes intersects the asymptotes at height b, makinga right triangle with base a and hypotenuse c. Thus,

c2 =a2 +b2. (1.138)

The roles ofc and a have interchanged in going from the ellipse to the hy-perbola.

A Reminder: Zenith and Flight-Path Angles

The angle between the velocity vector, v, and the local vertical is calledthe zenith angle, . The angle between the velocity vector and the localhorizontal is the flight-path angle, . The two angles are complements.

More on Newtons Cannon

Consider, again, Newtons cannon on a mountain top. Let it be suppliedwith an inexhaustible supply of powder, cannonballs, and all other necessaryequipment. Aim the cannon horizontally, so that the zenith angle is 90o

and the flight-path angle is zero. The magnitude of the angular momentumis h = rv sin = rv cos = rv. There are sub-orbital trajectories for thecannonball that are ellipses with the center of the Earth being at the focusfarther from the cannon, so the cannonball starts at apoapsis (or apogee).These orbits may be thought of as ballistic missile trajectories, for theyintersect the Earth. They are mostly interior to the circular orbit. Weimagine a first shot with very little powder that travels only a short distance,and successive shots with successively more powder until a circular orbitis obtained. More powder leads to more velocity, v, more specific angularmomentum, h, more specific energy,E, larger semi-latus rectum, p, largersemi-major axis,a, smaller eccentricity,e, and less semi-distance between the

foci, c. The circle may be thought of as the limiting case between interiorelliptical orbits and exterior ones. Compared with the interior orbits thecircular orbit has the maximum values ofv, h, E, a,and p and the minimumvalues of e and c. For all of these interior elliptical orbits the cannonballstarts at apoapsis (or apogee).


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Loading even more powder would lead to successively larger exterior or-

bits. The cannonball would start at periapsis (or perigee). The circularorbit would represent a limiting case to the exterior orbits and would havethe minimum values ofv, h,E,a,p,e, and c. Using even more powder wouldeventually result in a parabolic orbit and a family of hyperbolic orbits. Thecircular orbit would remain the minimum case for this entire family of orbits.

Degenerate Conics

Given the definition of a conic section as the intersection of a plane witha cone, I feel obliged to point out that such intersections include a singlepoint at the apex, which is also the force center, a single line that includes

the force center, and pairs of lines that intersect at the force center. Froma practical sense these orbits have little value to space travel because oftheir intersection with the force center. The point may be interpreted as astatic object at the force center. The line orbits could be interesting if wecould have a straight-line tunnel through the Earth that includes the Earthscenter. The straight-line orbit could also be terrifying if we identify an objectthat is on such an orbit and we live on the force center.

Problem 7 A tunnel through the EarthAssume the Earth to be spherical. Imagine a tunnel through the Earths

diameter, which, by definition, includes the Earths center. Hypothetically,

it would be possible to jump into such a tunnel and get free transportationto the other side of the Earth, courtesy of gravity. What would be the valuesofE, h,p,e,and a for such an orbit? What kind of conic would it be? Howlong would it take to get to the other side of the Earth?

To complete this problem it helps to think of the Earth as made of con-centric spherical shells. The shells outside of your current position in thetunnel exert a net gravitational force of zero - only the interior shells matter.Neglect air resistance in the tunnel and assume that the Earth has uniformdenisty.

1.6.4 What is B?

The vector constant of integration is at best unfamiliar. It is obviouslyimportant because it provides the reference direction from which the angle


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is measured, as shown in Figure 1.5-1 in the text. Our derivation of the

conic sections gives us a hint of the direction of B, for in those derivationsthe angle was measured from the horizontal. Going back a few steps, we

can solve the vector form of the trajectory equation to get

B= r h rr

. (1.139)

My first inclination is to determine B for the simplest case, which seems tobe a circle. In this case r andv are always perpendicular, so the magnitudeofhis rv and the cross product rhhas magnitude rv2 and its direction isalways outward from the center of the circular orbit. The direction ofrris always inward, so we only have to determine its magnitude relative torv

2

.The vector rr is a unit vector, so we simply need to know the magnitude of

, which isGM. We already know that for the circular orbit vcs=

rcs

, so

rcsv2cs = rcs

rcs=, (1.140)

which gives B = 0! We have chosen the special case in which B is zero.In retrospect, this makes sense, because a circle has no obvious referencedirection.

Next consider an elliptical orbit external to the circle, and specifically

examine the orbit at pericenter. Herer andv are still perpendicular, and ris still perpendicular toh, but v > vcs at that radius, so B points outward,away from the focus. Thus, B points in the direction from the focus topericenter.

Finally, consider an elliptical orbit internal to the circle. In this case thevector reverses because v < vcs, but the far focus is occupied, so the netagain points toward pericenter.

Problem 8 More on the direction of BConvince yourself that Bis toward pericenter for parabolic and hyperbolic

orbits.

In many advanced mechanics books B is shown to be closely related tothe Laplace-Runge-Lenz vector, specifically, B is that vector divided by themass, so we might give B the clumsy name specific Laplace-Runge-Lenz


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vector. Clearly, B is an additional constant of the motion. B depends on r,

as doesE, and it depends onh, so it does not add three vector componentsthat are all constants of the motion, only one. For more on this vector

see page 103 ofClassical Mechanics, third edition, by Goldstein, Poole, andSafko, published by Addison Wesley, copyright 2002.

1.6.5 The Eccentricity Vector

Following the above, we define

e= B/, (1.141)

the eccentricity vector, which points toward pericenter and has magnitude

equal to the eccentricity. This vector will simplify calculation of the orbitalelements in Chapter 2 and beyond. Working from the definition of Bwe find

e= B=v h rr

=v (r v) rr

=v2r (v r)v rr

, (1.142)

where the last step makes use of the BACCAB rule. This result will beuseful in future calculations.

1.7 Canonical Units

The text has been superseded since it was published in 1971. Now, astro-nomical distances and masses arewell known. Nonetheless, canonical unitsremain in use for convenience, and because every discipline uses units appro-priate to the problems that it investigates. In many cases these are not SIunits, and canonical units are not.

Please refer back to the problem titled The Mass of the Sun. In theyears after World War II, when captured German rocket technology madeit clear that space travel would occur, the mean radius of the Earths orbit- the Astronomical Unit - was not known with acceptable certainty, so themass of the Sun was not certain enough for detailed space-mission planning.

Planning was done in canonical units, and this continues because of theirsimplicity.The mathematical form of Keplers Third Law says

T2 =42

a3. (1.143)


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1.7. CANONICAL UNITS 35

IfG is certain andM is uncertain, but Solar orbits with great relative accu-

racy must be calculated for mission planning, then choose a system of unitsin which = GMcan be well known. Choose the Solar mass to be exactly1 mass unit, and choose the length of measuring sticks and the rate of clocksso thatG can also be exactly 1. Then, whenever the Solar mass is accuratelyknown all masses may be scaled to the accurate value, and the measuringsticks and clock rates can be similarly re-scaled. We note that having = 1will greatly simplify the calculations ofEfrom the energy equation, r fromthe equation of motion, and many other quantities describing orbits, such as

p ande. Solving the above equation for we find

= 42a3

T2. (1.144)

The leading term of 42 is a dimensionless constant, so in any system of units must have the dimensions of length cubed over time squared, matchingKeplers Third Law.

Choose the distance unit,DU, and the time unit, T U, so that = 1DU3

TU2 .To do this, first choose a circular reference orbit. For heliocentric orbitsthis is a circular orbit with a radius of one astronomical unit, or 1 AU, soDU = 1AU. Now choose a time unit that makes everything tie together asdesired. To do this, choose a time unit that gives the object in the referenceorbit unit speed, so DU/T U = 1. We already have the ability to calculate

the orbital speed as

v=2DU

T , (1.145)

where 2DU is the circumference of the circular orbit and T is its period.The two velocity definitions become consistent if

T U= T

2. (1.146)

Lets derive the quantities in the text. The period of the Earths orbitabout the Sun,T, is 365.24 days, which works out to be 3 .15567

107 seconds,

so T U = T /2 = 5.02241 106

seconds, which agrees with the values inAppendix A of the text to four significant figures. The Astronomical Unit is1.4959965 108 kilometers, so the speed unit, DU/T U= 29.7862 kilometersper second, which also agrees acceptably with the text. Following scientificstyle we will write kilometers per second as km s1, and proceed similarly.


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The gravitational parameter is = DU3/T U2 = 1.327291011 km3 s2, stillconsistent with the text to four significant figures.For any other gravitating body the reference orbit is a circular orbit that

just grazes its mean surface. The distance unit is the radius of this circularorbit. The time unit is still chosen so that the speed of the object in thisorbit is 1DU/TU, and is 1DU3/T U2. For Earth orbit Appendix A ofour text shows that 1DU = 2.09256 107 feet = 3963.19 miles = 3443.92nautical miles = 6378.14 kilometers. Again we choose the time unit so thatT U=T /2, but what is T?

By the year 1800 the circumference of the Earth was reasonably wellknown from a survey done by J. B. J. Delambre and P. F. A. Mechain, andthe density of the Earth was well known from the work of Henry Cavendish.

Cavendishs work also enabled the calculation of the gravitational constant,G, although this was not done until 1873 by A. Cornu and J. B. Baille. By thetime the Earth data were recognized as needed they were already available,while the solar data still were not. From our study of the circular orbit as aconic section we know

T = 2

r3/2, (1.147)

butT U=T /2, so

T U= T

2 =

r3/2

. (1.148)

Thus,andT Ucan be calculated from first principles using G = 6.671011m3 kg1 s2, M = 5.98 1024 kg, and r = DU = 6, 378, 000 m, to get = 3.98866 1014 m3 s2 or 3.98866 105 km3 s2 and T U = 805.171s. The values of 3.986012 105 km3 s2 and 806.811 s in the text makeuse of newer values of G, M, and DU, but the results agree very closely.The associated speed unit is 7.9215 km s1, which also agrees well with thetabulated value of 7.90536 km s1.

It would be pleasing if things could stay so simple, but they cannot, forthe metric system devised by the French at the time of their revolution wasnot well accepted in the English-speaking world. The meter was defined as

107

times the distance from the equator to the pole at sea level. The English,and by historical association Americans, used the nautical mile, which is onearc minute of latitude. Consequently, we still use feet and nautical miles,somewhat to our peril.

Please see


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1.7. CANONICAL UNITS 37

www.sizes.com/units/meter.htm

www.sizes.com/units/mile nautical.htm and

www.cnn.com/TECH/space/9909/30/mars.metric.02for more on these topics.

As an example, lets work the problem on page 41 of the text.


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Example 1 A Cooked Up Example

A space object is observed at an altitude of 1.046284 107

feet abovethe Earth traveling at 2.593625 104 feet s1 with a flight-path angle of 0o.Using canonical units determineE,h,p,e,ra,andrp.

Appendix A shows that in Earth units 1DU= 2.092567257107 feet and1DU/TU= 25936.24764 feet s1. This means that the object is 0.5000000DUabove the surface of the Earth and is traveling with a speed of 1.000000DU/TUon a flight-path angle of zero. The orbital radius is then the radius of theEarth plus 0.5000000 times the radius of the Earth, or exactly 1.5DU incanonical units. We calculate that

E=

v2

2

r

=1

2 1

1.5

=1

22

3

=

1

6

=

0.16666DU2/T U2. (1.149)

Similarly,h= rv cos = 1.5DU2/TU. (1.150)

The parameter is

p=h2

=

2.25DU4/T U2

1DU3/T U2 = 2.25DU. (1.151)

The eccentricity is

e=

1 +2Eh2

2 =

1 +0.3333 2.25

1 =

13

4=

14

=12

. (1.152)

The radius at apogee is

ra= p

1 e =2.25

0.5 = 4.5DU. (1.153)

The radius at perigee is

rp= p

1 + e=

2.25

1.5 = 1.5DU. (1.154)

I choose to give the authors a lot of credit for a clear example, but oth-erwise this has some problems with error bars and significant figures.


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Chapter 2

Orbit Determination from

Observations

The existence of Keplers Laws indicates to us that Kepler determined orbits,and Newton published a method in the Principia. This chapter develops themodern approach to orbit determination.

2.1 Coordinate Systems

We seek an inertial coordinate system in which to apply the results of Chapter1 before we can proceed. To specify a coordinate system we must give thelocation of the origin, the orientation of the fundamental plane (the X Yplane), the principal direction within the fundamental plane (the direction ofX), and the direction of the Z axis. Right-handed coordinates are assumed.There may be more than one such coordinate system, so we will also developthe mathematics for transforming from one system to another.

In reality none of the coordinate systems that we will define are trulyinertial, because all are accelerating to some degree. In many cases theaccelerations are small enough that the coordinates are good enough forpractical purposes.

We will start with four coordinate systems and add more when we needthem. The four systems are the Heliocentric-Ecliptic coordinate system, theGeocentric-Equatorial coordinate system, the Right Ascension-Declinationcoordinate system, and the Perifocal coordinate system. With experience itbecomes clear that the names at least hint at the location of the origin and

39


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40 CHAPTER 2. ORBIT DETERMINATION FROM OBSERVATIONS

the orientation of the fundamental plane for each system.

The Celestial Sphere is important to most celestial coordinate systems.This is a sphere of essentially infinite diameter onto which the stars, which aretaken to be at infinite distance, are projected. Thus, the apparent locationsof the stars are taken to be independent of the location of the observer onthe Earth. Objects of the Solar System and orbiting bodies are also seenprojected onto the Celestial Sphere, but because these objects are close tothe Earth their apparent locations on the Celestial Sphere depend on thelocation of the observer.

2.1.1 Heliocentric-Ecliptic Coordinates

As the name implies, Heliocentric-Ecliptic coordinates have their origin atthe center of the Sun. They are most useful for describing orbits around theSun. Their fundamental plane is the plane of the ecliptic, which is the planeof the Earths orbit. This is an infinite plane, extending to and intersectingthe Celestial Sphere. Within the plane of the ecliptic we can begin to definethe fundamental directions by determining the line of intersection betweenthe ecliptic plane and the Earths equatorial plane. A line from the centerof the Earth to the center of the Sun at the start of Northern Hemispherespring defines the fundamental direction along this line, and its intersection

with the Celestial Sphere is the First Point of Aries, also called the VernalEquinox. This direction is labeled X. TheZdirection is along the directionof the Earths orbital angular momentum vector. This puts the Y directiontoward the line from the center of the Sun to the center of the Earth at thestart of winter.

The mass distribution of the Earth is not perfectly uniform, so these di-rections precess slowly from torques applied primarily by the Sun and Moon,and their actual directions must be specified for a given year, called an epoch,i.e., epoch 2000. Today the First Point of Aries is actually in the constella-tion Pisces. Now you know why an astronomer is teaching orbital mechanics

to engineers.A closely related coordinate system is the Solar System Barycentric co-

ordinate system. This uses the same fundamental directions, but its originis at the center of mass of the Solar System, which is near the surface of theSun in the direction of Jupiter.


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2.1. COORDINATE SYSTEMS 41

2.1.2 Geocentric-Equatorial Coordinates

These coordinates have their origin at the center of the Earth. They areused to describe orbits around the Earth. The fundamental plane is that ofthe Earths equator, which is extended to intersect the Celestial Sphere atthe Celestial Equator. These coordinatesdo notrotate with the Earth. Thiswill force us to be keenly aware of our location on the Earth and the time,for Geocentric-Equatorial coordinates depend on them. TheXdirection istoward the Vernal Equinox, the same as X. The Zdirection points towardthe North Celectial Pole, near the North Star. There is an associated SouthCelestial Pole for the benefit of those in the Southern Hemisphere. I have seenit an it is real. The Ydirection is perpendicular to the X and Zdirections.

These directions are specified by unit vectors I , J, and K, respectively.

The Y and Z directions are not the same as those in the Geocentric-Equatorial coordinates because the rotational angular momentum vector ofthe Earth is inclined with respect to the orbital angular momentum vector.

2.1.3 Right Ascension-Declination Coordinates

These are closely related to Geocentric-Equatorial coordinates and use thesame fundamental directions. They are used to describe the coordinates ofstars and galaxies, which appear relatively fixed except for precession, andfor planets, which move because of their nearby location and orbital motion.Satellites, asteroids, and comets also appear to move in this system for thesame reasons of proximity and orbital motion. Their positions are oftendetermined by their locations relative to nearby stars, proving the utilityand importance of this system.

Right Ascension and Declination are both specified by angles - Right As-cension in the plane of the Celestial Equator eastward from Iand Declination

northward from the Celestial Equator.The origin of these coordinates may be at the center of the Earth or

somewhere on the surface of the Earth, called the topos. For nearby objectsthe coordinates depend on the location of the origin. These coordinates donot rotate with the Earth.


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42 CHAPTER 2. ORBIT DETERMINATION FROM OBSERVATIONS

2.1.4 Perifocal Coordinates

Perifocal coordinates can apply to any two-body gravitating system. Theorigin of Perifocal coordinates is the focus occupied by the central gravitat-ing body. The fundamental plane is the plane of the orbiting body, and thefundamental direction is the direction from the gravitating body to pericen-ter. It is veryeasy to describe an orbit in this system. The coordinate fromthe force center to pericenter is labeled x. The coordinate ninety degreesaway, in the plane of the orbit and in the direction of the motion of theorbiting body, is labeled y. The coordinate in the direction of the angularmomentum vector, h, is labeled z. The respective unit vectors are calledP , Q, and W. Pis a unit vector in the direction ofeand Wis a unit vecrot

in the direction ofh.

2.2 Classical Orbital Elements

Five independent quantities, called orbital elements are sufficient to describethe size, shape, and orientation of an orbit. One more is needed to locate theorbiting body at a particular place and time. The classical orbital elementsare:

a, the semi-major axis, a constant describing the size of the orbit,

e, the eccentricity, a constant defining the shape of the orbit,

i, the inclination, the angle between hand K,

, the longitude of the ascending node, an angle in the fundamental planebetween I and the ascending node of the line of nodes, measured counter-clockwise when viewed from the north side of the fundamental plane,

, the argument of periapsis, an angle in the orbital plane between the as-cending node and periapsis, measured in the direction of the satellites mo-tion, and

T, the time of periapsis passage.

Before going further, this is a good time to explain the naming of angles.If the measure of an angle starts from Ithen the angle is called a longitude.If the measure of an angle starts fromnthen the angle is called an argument.

There are important cases in which some of the quantities are not defined.For instance, in a circular orbit there is no periapsis, so is not defined. In


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2.2. CLASSICAL ORBITAL ELEMENTS 43

an orbit of zero inclination the fundamental and orbital planes coincide, so

there is no ascending node and is undefined. This coincidence is also calledan equatorial orbit.This list is sufficient, but not exhaustive. We saw in Chapter 1 that there

is a relation among a, e, and p, so any two can be used to determine thethird. Thus,p is often substituted for a. Modern radar observations make iteasier to find p than a for most Solar System objects.

There is also a good substitute for , the argument of periapsis, whichis , the longitude of periapsis. This is the angle fromI measured in thefundamental plane eastward to the ascending node plus the angle from thereto periapsis in the orbital plane. This is a very unusual physical quantitybecause it is the sum of two angles in different planes. If both and are

defined then = + . If is not defined then all is not lost, it still may bepossible to define . As long as there is a periapsis, meaning that the orbitis not circular, is the angle between I ande.

It is also possible to substitute for the time of periapsis passage, usually byspecifying the location of the satellite at the time of observation,to, also calledthe epoch. The true anomaly at epoch,o, is the angle in the orbital planebetween periapsis and the position of the satellite at to. This is the anglethat we called in deriving the trajectory equation, so it may be familiar.The subscript is meant to indicate that the angle has been observed at theparticular timeto. The argument of latitude at epoch,uo, is the angle in theorbital plane between the ascending node and the location of the satellite atto. For nonequatorial orbitsuo= +o. For equatorial orbitsuois undefined.The true longitude at epoch, o, is the angle from I measured eastward inthe fundamental plane to the ascending node plus the angle measured in theorbital plane, in the direction of motion, to the location of th

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